What is the integral from 0 to 4 of e^{-x} ?

The integral from 0 to 4 of e^{-x} is 1-1/(e^4)

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integral from 0 to 4 of e^{-x}

\int_{0}^{4} e^{- x} d x

1 - \frac{1}{e ^{4}}

Solution steps

\int_{0}^{4} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 4} - e^{u} d u

= \int_{0}^{- 4} - e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 4}^{0} - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{- 4}^{0} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- [e^{u}]_{- 4}^{0})

Simplify

= [e^{u}]_{- 4}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{4}}

= 1 - \frac{1}{e ^{4}}

n = 1 \sum \infty (\frac{1}{4})^{n}

f (x) = x^{4} + 5 x

\frac{d}{d x} (ln (1 + x) - x)

\int 8 w e^{7 w} d w

\int \frac{8 - 128 x}{1 - 64 x ^{2}} d x

What is the integral from 0 to 4 of e^{-x} ?
The integral from 0 to 4 of e^{-x} is 1-1/(e^4)